The admissible exponents of Strichartz estimate The Strichartz estimate for the Schrodinger equation is well-known:
$$
\|e^{it\Delta} u_0\|_{L^p_t L^q_x} \le C \|u_0\|_{L^2_x},
$$
where the exponent $p,q \in [2,\infty]$ satisfy
$$
\frac{2}{p}+\frac{d}{q} = \frac{d}{2}, \quad (p,q,d) \neq (2,\infty,2).
$$
I wonder whether we can replace $L^2$-norm of the R.H.S. to $L^r$-norm.
More precisely, I would like to know if we get an estimate like the following:
$$
\|e^{it\Delta} u_0\|_{L^p_t L^q_x} \le C \|u_0\|_{L^r_x}.
$$
The above estimate is invariant with respect to the scaling when $(p,q,r)$ satisfies
$$
\frac{2}{p}+\frac{d}{q} = \frac{d}{r}.
$$
 A: As I mentioned in comments, the answer is negative, even though I do not have a direct explanation. My explanation is indirect, going through the harmonic-analytic "restriction problem" of Stein.
Let me assume $p=q$ for simplicity. The estimate reads
$$\tag{1}
\lVert e^{it\Delta}u_0\rVert_{L^{2+4/d}(\mathbb R^{d+1})}\le C\lVert u_0\rVert_{L^2(\mathbb R^d)}.$$
Now define the measure $\mu(\tau, \xi):=\delta(\tau-\lvert\xi\rvert^2)$ in $\mathbb R^{d+1}$. The operator in the left-hand side reads
$$
e^{it\Delta}u_0(x)=\int_{\mathbb R^{d+1}} e^{i(t, x)\cdot (\tau, \xi)} f(\tau, \xi)\, d\mu(\tau, \xi), \quad f(\tau, \xi):=\widehat{u_0}(\xi).$$
On the other hand, using Plancherel, the norm in the right-hand side reads
$$
\lVert u_0\rVert_{L^2(\mathbb R^d)}^2=\lVert \widehat{u_0}\rVert_{L^2(\mathbb R^d)}^2=\int_{\mathbb R^{d+1}} f(\tau, \xi)\, d\mu(\tau, \xi).$$
This means that (1) is exactly the same as
$$\tag{2}
\lVert \widehat{f\mu}\rVert_{L^{2+4/d}(\mathbb R^{d+1})}\le C\lVert f\rVert_{L^2(d\mu)}, $$
which is the celebrated "Fourier adjoint restriction estimate" to the paraboloid $\{(\tau, \xi)\ :\ \tau=\lvert \xi \rvert^2\}$ of Stein--Tomas. This is also widely known as a "Fourier extension estimate".
It is (2) that does have a $L^p$ version. Indeed, it is conjectured that
$$
\tag{3}
\lVert \widehat{f\mu}\rVert_{L^q(\mathbb R^{d+1})}\le C_{p,r}\lVert f\rVert_{L^p(d\mu)}, $$
where $q>\frac{2(d+1)}{d}$ and $1/p=(d+2)/(qd)-2/d$ (scaling). If you express (3) in "PDE notation", by going backwards in the formulas of this answer, you obtain
$$
\tag{4}
\lVert e^{it\Delta}u_0\rVert_{L^{q}(\mathbb R^{d+1})}\le C_{p,d}\lVert \widehat{u_0}\rVert_{L^p(\mathbb R^d)},$$
which is what I meant in my comment to the question.

For reference on the Fourier restriction and extension (adjoint restriction) estimates, see the notes of Terry Tao.
